Chain Rule

Introduction

It is relatively easy to calculate the derivatives of simple functions, like polynomials or trigonometric functions.

But, what about more complicated functions?

For example,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\sin(3x)}   or  $g(x)=(x+1)^{8}?$

Well, the key to calculating the derivatives of these functions is to recognize that these functions are compositions.

For  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\sin(3x),}   it is the composition of the function  $y=3x$  with  $y=\sin(x).$

Similarly, for  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=(x+1)^{8},}   it is the composition of  $y=x+1$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=x^{8}.}

So, how do we take the derivative of compositions?

The answer to this question is exactly the Chain Rule.

Chain Rule

Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=f(u)}   be a differentiable function of  $u$  and let  $u=g(x)$  be a differentiable function of  $x.$ 

Then,  $y=f(g(x))$  is a differentiable function of  $x$  and

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y'=f'(g(x))\cdot g'(x).}

Warm-Up

Calculate  $h'(x).$

1)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)=\sin(3x)}

Solution:
Let  $f(x)=\sin(x)$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=3x.}
Then,  $f'(x)=\cos(x)$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g'(x)=3.}
Now,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)=f(g(x)).}
Using the Chain Rule, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {f'(g(x))\cdot g'(x)}\\&&\\&=&\displaystyle {\cos(3x)\cdot 3}\\&&\\&=&\displaystyle {3\cos(3x).}\end{array}}}

Final Answer:
$h'(x)=3\cos(3x)$

2)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)=(x+1)^{8}}

Solution:
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=x^{8}}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=x+1.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)=8x^{7}}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g'(x)=1.}
Now,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)=f(g(x)).}
Using the Chain Rule, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {f'(g(x))\cdot g'(x)}\\&&\\&=&\displaystyle {8(x+1)^{7}\cdot 1}\\&&\\&=&\displaystyle {8(x+1)^{7}.}\end{array}}}

Final Answer:
$h'(x)=8(x+1)^{7}.$

3)   $h(x)=\ln(x^{2})$

Solution:
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\ln(x)}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=x^{2}.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)={\frac {1}{x}}}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g'(x)=2x.}
Now,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)=f(g(x)).}
Using the Chain Rule, we have
${\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {f'(g(x))\cdot g'(x)}\\&&\\&=&\displaystyle {{\frac {1}{x^{2}}}\cdot 2x}\\&&\\&=&\displaystyle {{\frac {2}{x}}.}\end{array}}$

Final Answer:
$h'(x)={\frac {2}{x}}$

Exercise 1

Calculate the derivative of  $h(x)=(\sin x+\cos x)^{4}.$

Using the Chain Rule, we have

${\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {4(\sin x+\cos x)^{3}(\sin x+\cos x)'}\\&&\\&=&\displaystyle {4(\sin x+\cos x)^{3}((\sin x)'+(\cos x)')}\\&&\\&=&\displaystyle {4(\sin x+\cos x)^{3}(\cos x-\sin x)}.\end{array}}$

So, we have

$h'(x)=4(\sin x+\cos x)^{3}(\cos x-\sin x).$

Exercise 2

Calculate the derivative of  $h(x)=\sin ^{3}(2x^{2}+x+1).$

First, notice  $h(x)=(\sin(2x^{2}+x+1))^{3}.$

Using the Chain Rule, we have

$h'(x)=3(\sin(2x^{2}+x+1))^{2}\cdot (\sin(2x^{2}+x+1))'.$

Now, we need to use the Chain Rule a second time. So, we get

${\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {3(\sin(2x^{2}+x+1))^{2}\cos(2x^{2}+x+1)\cdot (2x^{2}+x+1)'}\\&&\\&=&\displaystyle {3\sin ^{2}(2x^{2}+x+1)\cos(2x^{2}+x+1)(4x+1).}\end{array}}$

So, we have

$h'(x)=3\sin ^{2}(2x^{2}+x+1)\cos(2x^{2}+x+1)(4x+1).$

Exercise 3

Calculate the derivative of  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h(x)={\frac {x^{2}\sin x+1}{x^{2}\cos x+3}}.}

Using the Quotient Rule, we have

$h'(x)={\frac {(x^{2}\cos x+3)(x^{2}\sin x+1)'-(x^{2}\sin x+1)(x^{2}\cos x+3)'}{(x^{2}\cos x+3)^{2}}}.$

Now, we need to use the Product Rule. So, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {(x^{2}\cos x+3)(x^{2}(\sin x)'+(x^{2})'\sin x)-(x^{2}\sin x+1)(x^{2}(\cos x)'+(x^{2})'\cos x)}{(x^{2}\cos x+3)^{2}}}\\&&\\&=&\displaystyle {{\frac {(x^{2}\cos x+3)(x^{2}\cos x+2x\sin x)-(x^{2}\sin x+1)(-x^{2}\sin x+2x\cos x)}{(x^{2}\cos x+3)^{2}}}.}\end{array}}}

So, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle h'(x)={\frac {(x^{2}\cos x+3)(x^{2}\cos x+2x\sin x)-(x^{2}\sin x+1)(-x^{2}\sin x+2x\cos x)}{(x^{2}\cos x+3)^{2}}}.}

Exercise 4

Calculate the derivative of  $f(x)={\frac {e^{x}}{x^{2}\sin x}}.$

First, using the Quotient Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.} \end{array}}

Now, we need to use the Product Rule. So, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}$

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$